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C1 - coordinate geometry

How do I do part d? Any hints on how to start?
Thanks :smile:
c1 - Coordinate Geometry.jpg46.9KB

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Reply 1
Avatar for zed963
zed963
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Original post by Super199
How do I do part d? Any hints on how to start?
Thanks :smile:


Just sub p into the equation, if it is = to 0 then it satisfies.
Reply 2
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Super199
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Original post by zed963
Just sub p into the equation, if it is = to 0 then it satisfies.

What equation?
Reply 3
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zed963
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Original post by Super199
What equation?


p^2-4p-16=0
Reply 4
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Super199
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Original post by zed963
p^2-4p-16=0

I don't know the value of p :s-smilie:.
I am given AC. Where A is (2,5) and c is (p, some y coordinate) which is equal to 5? It's worth 4 marks so I dunno
Reply 5
Avatar for zed963
zed963
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Original post by Super199
I don't know the value of p :s-smilie:.
I am given AC. Where A is (2,5) and c is (p, some y coordinate) which is equal to 5? It's worth 4 marks so I dunno


It clearly says " has x coordinate which is equal to p"

Post your working out to the question.
(edited 10 years ago)
Reply 6
Avatar for james22
james22
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What is the coordinates of point C in terms of p?

What is the length of AC in terms of p?

Answer those and you should be able to solve the rest.
Reply 7
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james22
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Original post by zed963
Just sub p into the equation, if it is = to 0 then it satisfies.


You cannot do this, you don't know what p is yet.
Reply 8
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zed963
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Original post by james22
You cannot do this, you don't know what p is yet.


I assumed the OP already had P.
Reply 9
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Super199
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Original post by james22
What is the coordinates of point C in terms of p?

What is the length of AC in terms of p?

Answer those and you should be able to solve the rest.

If I could work out the y coordinate of p I should be able to do this. I can't seem to work it out.
Then for AC I would say
AC2=(p−2)2+(ycoordinate−5)2AC^2=(p-2)^2+(y coordinate -5)^2?
Reply 10
Avatar for zed963
zed963
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Original post by Super199
If I could work out the y coordinate of p I should be able to do this. I can't seem to work it out.
Then for AC I would say
AC2=(p−2)2+(ycoordinate−5)2AC^2=(p-2)^2+(y coordinate -5)^2?


ImageUploadedByStudent Room1397673804.420931.jpg

I've tried doing it but its not as easy as it looks.


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Reply 11
Avatar for davros
davros
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Original post by Super199
If I could work out the y coordinate of p I should be able to do this. I can't seem to work it out.
Then for AC I would say
AC2=(p−2)2+(ycoordinate−5)2AC^2=(p-2)^2+(y coordinate -5)^2?


You know what the y coordinate will be in terms of p because C lies on the line and you worked out the equation of the line earlier in the question.
Reply 12
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Super199
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Original post by davros
You know what the y coordinate will be in terms of p because C lies on the line and you worked out the equation of the line earlier in the question.

Oh wow yh hang on I will have another go. Thanks :smile:
I have the solution to this? Want hints?
C= (P,-1/2P+6)
You then plug this into the AC^2 formula
Reply 16
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james22
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Original post by zed963
I assumed the OP already had P.


RTFQ
Reply 17
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zed963
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Original post by Galileo Galilei
C= (P,-1/2P+6)


Can you explain where you've got this from?
Reply 18
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zed963
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Original post by james22
RTFQ


Sorry guys.
Original post by zed963
Can you explain where you've got this from?


Mathematical logic

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